tissue mechanics: mechanical properties of bone, muscle, tendon, etc
DESCRIPTION
Overall course plan. Tissue mechanics: mechanical properties of bone, muscle, tendon, etc. Kinematics: quantification of motion, with no regard for the forces that govern motion Kinetics: forces in human movement. Kinematics. 1 -D linear kinematics Angular kinematics - PowerPoint PPT PresentationTRANSCRIPT
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• Tissue mechanics: mechanical properties of bone, muscle, tendon, etc.
• Kinematics: quantification of motion, with no regard for the forces that govern motion
• Kinetics: forces in human movement
Overall course plan
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion
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Reading
• Linear: Enoka chapter 1 pp. 3-23• Loco: Enoka chapter 4 pp. 141-146• Angular: Enoka chapter1 pp. 23-29
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion
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• 60-6000 images/sec• Put markers on
anatomical landmarks• Computer tracks
marker position
Video Kinematics
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Kinematics example video
• http://www.vicon.com/applications/life_sciences.html
• http://www.youtube.com/watch?v=USOYUMN5nwU
• http://www.youtube.com/watch?v=IJ4tndpwL-o&mode=related&search=
• http://www.youtube.com/watch?v=frNpkIH8_vo&mode=related&search=
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Kinematics• 1D = movement in one direction.
– e.g just the forward movement of a human runner during a sprint
• 2D = movement in a plane, components in 2 directions– e.g. side view of hip of a person walking
• 3D = movement anywhere in space– an owl swooping to catch jackrabbit, knee movements
• Angular = movements that include rotation– e.g., a biceps curl, alligator biting, figure skater
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1 dimensional linear kinematics
• Measurement Rules• Definitions of terms used in kinematics• Equations and examples for constant
acceleration motion• 1D Kinematics example
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Units
• SI system– Length : meters (m)– Mass : kilograms (kg)– Time : seconds (s)
• Other units– Angle : radians (rad)
• 2p radians = 360o
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Prefixes and Changing Units
• Giga (G) 1x109 • Mega (M) 1x106
• Kilo (k) 1x103
• Deci (d) 1x10-1
• Centi (c) 1x10-2
• Milli (m) 1x10-3
When changing units, always be sure to CANCEL units appropriately!
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1 dimensional linear kinematics
• Measurement Rules• Definitions of terms used in kinematics• Equations and examples for constant
acceleration motion• 1D Kinematics example
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Kinematics terms: position• position (r) = location in
space• Units : meters (m)• Must be defined with
respect to a baseline value or axis
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Kinematics terms: Displacement• Displacement = Change in position
= ∆r = (rf - ri)
• rf = final position; ri = initial position• Units = meters (m)• Involves space and time
Positi
on r,
(m)
Time (s)
ri
rf
ti tf
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Kinematics terms: Velocity (v)• rate of change of position• vave = ∆r / ∆t
• vinstantaneous = dr / dt (slope of position vs. time)• Units = m / s
Velo
city
v, (m
/s)
Time (s)
vi
ti tf
Positi
on r,
(m)
Time (s)
ri
ti tf
rf
vf
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Kinematics terms: acceleration (a)
• rate of change of velocity• aave = ∆v / ∆t
• ainstantaneous = dv / dt = slope of velocity vs. time• units = m / s2
Velo
city
v, (m
/s)
Time (s)
vi
ti tf
Positi
on r,
(m)
Time (s)
ri
ti tf
rf
vf
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1 dimensional linear kinematics
• Measurement Rules• Definitions of terms used in kinematics• Equations and examples for constant
acceleration motion• 1D Kinematics example
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Special case: Constant acceleration• Example: Projectile motion when air resistance is
negligible• Gravitational acceleration = 9.81 m / s2
• in the downward direction!!!• g = 9.81 m/ s2
• Useful for analyzing a jump (frogs, athletes), the aerial phase of running, falling
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Projectile Motion
• The only significant force that the object experiences while in the air will be that due to gravity
• Flight time depends on vertical velocity at release and the height of release above landing surface
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Symmetry in projectile motion
• A comparison of the kinematics at the beginning and end of the flight reveals: vi = - vf & vi
2 = vf2
ri = rf & rf - ri = 0
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Equation 1 for constant acceleration
• Expression for final velocity (vf) based on initial velocity (vi), acceleration (a) and time (t)
vf = vi + at
• If vi = 0vf = at
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Problem solving strategy• Define your coordinate system!• List all known variables (with UNITS!)• Write down which variable you need to know
– (with UNITS!)• Figure out which equations allow you to solve for
the unknown variable from the known variables.
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Example using equation 1• An animal jumps vertically. The animal’s takeoff
or initial velocity is 3 m/s. How long does it take to reach the highest point of the jump? (Air resistance negligible).
A) - 0.3 sB) 0.03 sC) - 0.03 sD) 0.3 s
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Equation 2 for constant acceleration• Expression for change in position (rf - ri) in terms
of initial velocity (vi), final velocity (vf), acceleration (a) and time (t).
rf - ri = vit + ½ at2
• If vi = 0, thenrf - ri = ½ at2
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Example using equation 2What was the net jump height of the animal in
example #1. In other words, how high did it jump once its feet left the ground?
A) 46 mB) 46 cmC) 1.34 mD) -1.34 m
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Equation 3 for constant acceleration
• An expression for the final velocity (vf) in terms of the inital velocity (vi), acceleration (a) and the distance travelled (rf - ri).
vf2 = vi
2 + 2a (rf - ri)
• If vi = 0, thenvf
2 = 2a (rf - ri)
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Example using Equation 3
A runner starts at a standstill and acceleratesuniformly at 1 m/s2. How far has she travelled at
the point when she reaches her maximum speed of 10 m/s?
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Example using Equation 3
A runner starts at a standstill and accelerates uniformly at 1 m/s2. How far has she travelled at the point when she reaches her maximum speed of 10 m/s?A) 5 mB) 10 mC) 50 mD) 100 m
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1 dimensional linear kinematics
• Measurement Rules• Definitions of terms used in kinematics• Equations and examples for constant
acceleration motion• 1D Kinematics example
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1-D kinematic analysis of a 100 meter race
• Question to answer: How long does it take to reach top speed?
• Videotape the sprint• Measure the position of a hip marker in each
frame of video
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100 meter sprint
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100 meter sprint
vmax reached at 6 seconds, at 50 meters.
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100 meter sprint
vmax = 10 m/s aave = 1.66 m/s2
(0 - 6 seconds)
Acceleration (m / s2)
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion
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Position• Linear
symbol = runit = meter
• Angularsymbol = unit = radians or degrees6.28 rad = 360°
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Joint angles
• Ankle: foot - shank• Knee: shank - thigh• Hip: thigh - trunk• Wrist• Elbow• ShoulderKnee
Hip
Ankle
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Knee
Hip
Ankle
• Measured relative to a fixed axis (e.g., horizontal)
Segment angles
Shank
Trunk
Trunk
Thigh
Shank
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Displacement• Linear
symbol = ∆r∆r = rf - ri
unit = meter• Angular
symbol = ∆∆ = f - i
unit = radian
∆ 1
2
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Angular ---> linear displacement• s = distance travelled (arc)• s = radius • ∆∆ must be in radians
∆
s
radius1
2
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Joint angle change: Flexion
• Flexion: decrease in angle between two adjacent body segments
• Bird’s eye view of arm on table
Elbow
Flexion
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Joint angle change: Extension
• Extension: increase in angle between two adjacent body segments
Extension
Elbow
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Kneeextension
Knee extension
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Knee FLEXION
Kneeflexion
knee
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Ankleext
A Squat Jump
Hipext
Kneeext
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EMG: Electromyography• Measures electrical activity of muscles
– indicates when a muscle is active
Ankleext
Hipext
Kneeext
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Time (s)
Hipangle
Kneeangle
Ankleangle
0.15 0.300
Hip extensor EMG
Knee extensor EMG
Ankle extensor EMG
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Alternative joint angle definition
Full extension: alt = 0 (degrees or rad)
alt
norm
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0
1
2
0
1
Run
Walk
Flexion
Time (% of stride)
Enoka 4.5Kneeangle(rad)
Flex-Ext Flex-Ext
SmallFlex-Ext Flex-Ext
Stance Swing
stance
stance
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Velocity• Linear– symbol = v– v = ∆r / ∆t– unit = meters per second
• Angular– symbol = = “omega””– = ∆ / ∆t– unit = radians per second– human body: we define
extension as positive
∆
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• v = r
Linear velocity (v) & angular velocity ()
v
r
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A person sits in a chair and does a knee extension. The knee angle changes by 0.5 radians in 0.5 seconds.
What is the magnitude of the angular velocity of the knee, the linear velocity of the foot and the linear
displacement of the foot? (Shank length is 0.5 meters)
A) 1 rad/s, 0.5 m/s, 0.25rad
B) 0.5 rad/s, 0.5 rad/s, 0.25rad
C) 1 rad/s, 0.5 m/s, 0.25 m
D) 0.25 rad/s, 0.5 m/s, 0.25m
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Angular acceleration• Linear
– symbol = a– a = ∆v /∆t– unit = m / s2
• Angular– symbol = – = “alpha”– = D / Dt– unit = rad / s2
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Arm Motion example
Muscles act as “motors” and “brakes”But they only pull
Displacement, velocity, accelerationdr/dt, dv/dt
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Time
(rad)
(rad/s)
(rad/s2)
E
F
E
F
E= extension, F = flexion
1
2
3
1 2 3E
F
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(rad)
(rad/s)
(rad/s2)
Triceps brachialis (ext)
Brachioradialis (flex)
Biceps brachialis (flex)
Flex
Ext
Enoka example1.8
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• Even if vi =vf, there is still a radial acceleration
• Even if i =f, there is still a radial acceleration
• aradial (m/s2): ∆ direction of linear velocity vector.
• aradial = r2 = v2 / r– ar is never zero during circular motion
Linear acceleration (a) in angular motion:radial and tangential acceleration
ar
vi
vf
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• a = aradial + atangential (vector sum)
• atangential (m/s2): causes ∆ & ∆v
• atangential = r = ∆v / ∆tat is zero if |v| or is constant
ar
at
Linear acceleration (a) in angular motion
v
v
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Total linear acceleration (a)
• a = at + ar (vector sum)
• |a| = (at2 + ar
2)0.5
• = tan-1(at / ar)
ar
aat
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A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What
were the magnitudes of the tangential & radial components of the acceleration?
A) at = 981 m/s2; ar = 2 m/s2
B) at = 0 m/s2; ar = 4 m/s2
C) at = 0 m/s2; ar = 2 m/s2
D) at = 0 m/s2; ar = 2 rad/s2
E) at = 981 m/s2; ar = 4 rad/s2
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A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What
were the magnitudes of the tangential & radial components of the acceleration?
at = r = 0; r = 2 meters at = r = (2 m)• (0 rad/s2) = 0 m/s2
ar = r2
= 1 rad/secar = (2 meters) • (1 rad/s)2 = 2 m/s2
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Linear & Angular Motion Equations
• On equation sheet• analogous Only for constant acceleration
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Make a list of examples in human movement where radial and tangential accelerations are important
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Angular Kinematics Example # 1
A hammer thrower releases the hammer after reaching an angular velocity of 14.9 rad/s. If the hammer is 1.6 m from the shoulder joint, what is the linear velocity of release?
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A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race.
a. How long does it take for her to run the 200 m?
b. What is her speed when she crosses the finish line? c. What is her angular velocity when she finishes? d. What was her angular acceleration during the first two seconds? e. What is her angular acceleration when she finishes? f. Determine her radial and tangential accelerations 1 second after the start of the race.
Angular Kinematics Example # 2
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion
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2-D linear kinematics• 1-D vs. 2-D kinematics• conventions• vectors• constant acceleration motion• applications
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1-D analysis: 100 meter sprint example
• We only considered the horizontal motion.
vv
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Hip moves vertically & horizontally.
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• Horizontal & vertical velocity components
vv
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2-D linear kinematics• 1-D vs. 2-D kinematics• conventions• vectors• constant acceleration motion• applications
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Running direction
Y
XX
Y
X
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Y
X
Z
X
Z
X
Z
Y
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Sign conventions• Y axis (vertical): Positive is upward• X axis (horizontal): Positive is in direction of
movement.
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2-D linear kinematics• 1-D vs. 2-D kinematics• conventions• vectors• constant acceleration motion• applications
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Vectors in kinematics• VECTOR: magnitude and direction.
– Length of vector indicates magnitude.– Direction of vector indicates direction of motion.
Large horizontalvelocity
Small verticalvelocity
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Which Vectors Are Equal?
A) A&BB) A,B&EC) C&ED) None of them are equal
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2-D kinematic analysis
• Vector analysis: divide movement into components & analyze separately
• Reason: different forces &accelerations act in each direction– vertical: gravitational acceleration– horizontal & lateral: no grav. accel.
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Vector components
• Right triangle ---> use trigonometry.• |v| = magnitude of vector
= angle of vector with the horizontal
• vx = horizontal component
vy = vertical component
|v| v
vx
vy
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• What are the horizontal (vx) and vertical (vy) components of the vector (v)?
• cos = vx / v thus, vx = v cos • sin = vy / v thus, vy = v sin
Determining vector components
v
vx
vy
|v|
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Example of determining components
• sin 60° = vy / v thus, vy = v * sin 60°
• vy = 2 * 0.87 = 1.74 m/sec
• cos 60° = vx / v thus, vx = v * cos 60°
• vx = 2 * 0.50 = 1 m/sec
60°
|v| = 2m/sec
vx
vy
60°
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How to find resultant vector
• Finding tan = vy / vx thus, = tan-1(vy / vx)
• Finding |v||v| = (vx 2 + vy
2)0.5
v = ?
vx
vy
=vx
vy
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Example: Find resultant vector
= tan-1(vy /vx) = tan-1(3 / 2) = 56.3° |v| = (vx
2 + vy2)0.5
|v| = (22 + 32)0.5 = (13)0.5 = 3.6m/s
|v|
vx = 2m/s
vy = 3m/s
|v| = ? = ?
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2-D kinematic analysis
• Vector analysis: divide movement into components & analyze separately
• Reason: different forces/accelerations act in each direction– vertical: gravitational acceleration– horizontal & lateral: no grav. accel.
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2-D linear kinematics• 1-D vs. 2-D kinematics• conventions• vectors• constant acceleration motion• applications
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Projectile Motion• The only significant force that the object
experiences while in the air will be that due to gravity
• Flight time depends on vertical velocity at release and the height of release above landing surface
• gravity causes the trajectory to deviate from a straight line into a parabola
• Because there is no force in the horizontal direction, the horizontal acceleration is zero
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Symmetry in projectile motion
• A comparison of the kinematics at the beginning and end of the flight reveals: vi = - vf & vi
2 = vf2
ri = rf & rf - ri = 0
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Example: A long jumper leaves the ground with |v| = 7.6 m/s at an angle of 23°. Which equation(s) do you need to
calculate how far she jumps?1. vf = vi + at2. rf - ri = vit + ½ at2
3. vf2 = vi
2 + 2a (rf - ri)
A) 1,2B) 2,3C) 1,2,3D) 2
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Step 1: Break the vector into components.
• vyi = vi * sin
• vyi = 7.6 * sin 23° = 3 m/s
• vxi = vi * cos
• vxi = 7.6 * cos 23° = 7 m/s
23°
7.6 m/s vi
vxi
vyi
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Step 2: From vyi, calculate how long the jumper stayed in the air.
• When air resistance is negligible, vyf = -vyi
• vyf = vyi + at (Eq. 1)
• -vyi = vyi -9.81t ---> 2vyi = 9.81t ---> t = vyi / 4.9• t = 3/4.9 = 0.6 seconds.
7.6 m/s
7 m/s
3 m/s = 23°
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• vx is constant throughout jump.
• ∆rx = vx * t
• ∆rx = 7 * 0.6 = 4.2 meters
Step 3: From the time in the air and vx, calculate the horizontal distance travelled while in the air.
7.6 m/s
7 m/s
3 m/s = 23°
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Angular/2-D Example # 1
A hammer thrower releases the hammer after reaching an angular velocity of 14.9 rad/s. If the hammer is 1.6 m from the shoulder joint, what is the linear velocity of release?
If the release angle is 30 degrees, and release height is 1.5 m, how far away did the hammer land?
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2-D ExampleA baseball is thrown with a velocity of 31 m/s at an angle of 40 degrees from a height of 1.8m. Calculate the:• vertical and horizontal velocity components• time to peak trajectory• height of trajectory from point of release• total height of parabola• time from peak to the ground• total flight time• range of the throw
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Equations for Constant Acceleration
vf = vi + atrf - ri = vit + ½ at2
vf2 = vi
2 + 2a (rf - ri)
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion
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Locomotion kinematics
• What defines gaits?• Components of a stride• Vertical displacement of the body• Horizontal velocity of the body• How SF and SL change with speed
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What defines a walk vs. a run?
• Kinematic• Kinetic
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Walking and running speeds• Walking speeds: 0.5 - 2 m/s
• Running speeds: 2.5 m/s - max – Peak for elite male athletes = 12 m/s
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Locomotion kinematics
• What defines gaits?• Components of a stride• Vertical displacement of the body• Horizontal velocity of the body
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Kinematic terms for gait analysis• Stride: One complete cycle from an event (e.g.,
right foot touch-down) to the next time that event occurs.
• Stance phase: Time when a limb is in contact with the ground.
• Swing phase: Time when a limb is NOT in contact with the ground.
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Kinematic terms for gait analysis• Double support: Time when 2 limbs are in
contact with the ground.• Aerial phase: Time when no feet are on the
ground.
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Walk
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Run
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Stance vs Swing phase
• In walking: 60%/40% : Stance/Swing• As speed increases:—absolute time spent in stance decreases—relative time spent in stance decreases—true for both walking and running
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Walking vs Running
Variable Walking Running Sprint
Speed (m/s) 0.67-1.32 1.65-4.0 8.0-9.0
Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50
Cadence (steps/min) 79 - 118 132-200
Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0
Cycle Time (s) 1.55 – 1.02 0.91- 0.73 0.57-0.5
Stance (% of gait cycle) 66 -60 59 - 30 25-20
Swing (% of gait cycle) 34 - 40 41 - 70 75- 80
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Comparison of walk vs. run
• stance time: run < walk• swing time: run > walk• stride time: run << walk
• aerial time: run >> walk (0)• double support: run (0) << walk
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To increase my walking speed my steps must get longer.
A)TrueB) FalseC) It depends
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Speed = SF * SL• Speed (m/s) = SF (strides/s) * SL (m/strides)
– stride frequency: number of strides per second– stride length: distance covered by one stride
• Can increase speed by increasing SF, SL or both• But step length can only be increased so much
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Speed = SF * SL• Speed (m/s) = SF (strides/s) * SL (m/strides)• Question: Do people increase speed by
increasing SF and/or SL during walking and running?
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Walking vs Running
Variable Walking Running Sprint
Speed (m/s) 0.67-1.32 1.65-4.0 8.0-9.0
Stride Length (m) 1.03 – 1.35 1.51 – 3.0 4.60 – 4.50
Cadence (steps/min) 79 - 118 132-200
Stride Rate (Hz) 0.65 – 0.98 1.10 – 1.38 1.75 – 2.0
Cycle Time (s) 1.55 – 1.02 0.91- 0.73 0.57-0.5
Stance (% of gait cycle) 66 -60 59 - 30 25-20
Swing (% of gait cycle) 34 - 40 41 - 70 75- 80
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Locomotion kinematics
• What defines gaits?• Components of a stride• Vertical displacement of the body• Horizontal velocity of the body
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Walk Run
• Hip highest atmid-stance.• Leg is straighter.
• Hip lowest atmid-stance.• Leg more bent.
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Walk
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Run
Time (s)
0.25 0.750.500
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Locomotion kinematics
• What defines gaits?• Components of a stride• Vertical displacement of the body• Horizontal velocity of the body
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Make a graph of the horizontal velocity vs. time for a walking stride
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Walk
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Run
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Walk and run• Forward velocity fluctuates, reaching its slowest
value at the middle of the stance phase.• Vx never reaches zero
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Walking vs. running• Double stance?• Aerial phase?• Position of hip:
– highest at mid-stance in running.– lowest at mid-stance in walking.
• Leg posture during stance phase.– straighter in walking than running
• Both walk & run: forward velocity reaches minimum at the middle of the stance phase.
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Kinematics
• 1-D linear kinematics• Angular kinematics• 2-D linear kinematics• Locomotion